When evaluating a Kriging prediction, it is possible to include a hyperparameter $\lambda$ to account for noise in the data. $\lambda$ can be estimated as a parameter in a maximum likelihood estimation (MLE) of the observed data. One effect is that the Kriging estimators will no longer pass through the sample points, and $\lambda$ can be seen as the variance of the noise of the samples' observed values.
Please correct me if I'm saying something wrong!
This approach is fine, but it assumes all the observables have the same error. What if I knew the error in some (all) of the observables? Would it be possible to manually add the error of the i-th sample $\sigma^2_i$ to the corresponding diagonal entry of the correlation matrix? i.e., use $\Psi_{ij} + \sigma_i^2 \delta_{ij}$ instead of simply $\Psi_{ij}$
If so, how could I calculate the variance of the Kriging estimators? My equation for the variance of the Kriging estimators for the simple $\lambda$ case is:
$$\hat{\sigma}^2 [ 1 + \lambda - \psi^T( \Psi + \lambda \bf{I})^{-1} \psi]$$
I don't understand the derivation of this equation, so I don't know how to change it to accomodate individual errors.
Additionally, I feel using individual errors would affect the calculation of the MLE when determining the hyperparameters: if an outlier has a large error, the resulting MLE distribution should have a lower $\hat{\sigma}$. (Right?)
Any insights would be greatly appreciated!